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Let $\Omega = [0,1]$, and let $\mathcal{F}$ be the Borel $\sigma$ field of $[0,1]$. Also let $\mathbb{P}$ denote the Lebesgue measure on $\Omega$ ($\mathbb{P}(\Omega) = 1$). Let $\mathcal{F}_n$ denote the smallest $\sigma-$field containing the dyadic intervals $J_i = [i2^{-n}, (i+1)2^{-n}]$ for all $0 \leq i \leq 2^n - 1$.

Suppose also that $f:[0,1] \rightarrow \mathbb{R}$ be such that $\mathbb{E}(|f|^2) < \infty$.

I define $f_n = \mathbb{E}(f \mid \mathcal{F}_n)$ and know that $\{f_n\}$ is an $L_2$-bounded martingale, and hence $f_n$ converges to some $g$ in $L_2$.

Given all this, I assume that $f$ is continuous, and am trying to show that $g(x)=f(x)$ for all $x$ with probability 1. It seems rather trivial, but am not sure how to proceed. Would anyone have any hints? thanks.

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Note that $f_n$ can be calculated explicitly using the following result:

A mapping $g: [0,1] \to \mathbb{R}$ is $\mathcal{F}_n$-measurable if, and only if, there exist constants $c_0,\ldots,c_{2^n-1}$ such that $$g(x) = \sum_{j=0}^{2^n-1} c_j 1_{[j 2^{-n},(j+1) 2^{-n}]}.$$

This implies that

$$\mathbb{E}(f \mid \mathcal{F}_n) = \sum_{j=0}^{2^n-1} c_j 1_{[j 2^{-n},(j+1) 2^{-n}]}$$

for suitable constants $c_j$. Using the properties of the conditional expectation, we find that

$$c_j = \int_{j 2^{-n}}^{(j+1) 2^{-n}} f(y) \, dy.$$

From this representation for $\mathbb{E}(f \mid \mathcal{F}_n)$ and the continuity of $f$ it follows that

$$\lim_{n \to \infty} f_n = f.$$

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  • $\begingroup$ saz, 1. do you need continuity? 2. Did you use $\mathcal L^2$? $\endgroup$ – BCLC Apr 6 '18 at 19:13
  • $\begingroup$ @BCLC 1) The continuity is needed to show that $f_n \to f$; see the last part of my answer. 2) The OP used the $L^2$-condition on $f$ to conclude that $(f_n)_n$ is an $L^2$-bounded martingale. $\endgroup$ – saz Apr 6 '18 at 19:26
  • $\begingroup$ But saz, you could come up with a different proof without using continuity? $\endgroup$ – BCLC Apr 6 '18 at 20:11
  • $\begingroup$ @BCLC Yeah, I guess; one could for instance apply Lévy's convergence theorem (for martingales)... as far as I can see it doesn't require continuity of $f$. $\endgroup$ – saz Apr 7 '18 at 6:21

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